Post on 20-May-2020
MECH 420: Finite Element Applications
Lecture 12: Grid Elements and Grid Analysis.
The grid element equations in terms of elemental coords:
3 2 3 2
11
1
2 21
23 2 3 2
2
2
2 2
12 0 6 12 0 6
ˆˆ 0 0 0 0ˆˆ
6 0 4 6 0 2ˆˆ
12 0 6 12 0 6ˆˆ 0 0 0 0
6 0 2 6 0 4
yy
x
z
y
x
z
EI EI EI EIL L L L
GJ GJ df L Lm EI EI EI EIm L L L L
EI EI EI EIfL L L Lm
GJ GJmL L
EI EI EI EIL L L L
⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−⎧ ⎫⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪
−⎢ ⎥⎪ ⎪⎪ ⎪ ⎢ ⎥=⎨ ⎬ ⎢ ⎥⎪ ⎪ − − −⎢ ⎥⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎢ ⎥−⎪ ⎪⎩ ⎭ ⎢ ⎥⎢ ⎥⎢ − ⎥⎣ ⎦
1
1
2
2
2
ˆˆ
ˆˆ
x
z
y
x
z
d
φφ
φφ
⎧ ⎫⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭
x
y
z
It is obvious that we need to apply global coordinates in solving grid problems.
2
3
MECH 420: Finite Element Applications
Lecture 12: Grid Elements and Grid Analysis.
The three-dimensional rotation matrix, .
The grid element, by convention, lies in the xz plane of the global frame.
This is defined in Logan’s steps 1 through 4 when we created ourgrid element.
R
ˆˆ ˆ( ) ( ) ( )ˆ ˆ ˆˆˆ ˆˆ ˆ ˆ( ) ( ) ( )ˆˆ ˆ ˆˆ ˆ( ) ( ) ( )
x x x
y y y
z z z
i i j i k ix x x c c c xy R y i j j j k j y c c c yz z z c c c zi k j k k k
θ ψ φθ ψ φθ ψ φ
⎡ ⎤⋅ ⋅ ⋅⎧ ⎫ ⎧ ⎫ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥= = ⋅ ⋅ ⋅ =⎢ ⎥⎨ ⎬ ⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥⋅ ⋅ ⋅⎩ ⎭ ⎩ ⎭ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭⎢ ⎥⎣ ⎦
ˆ, , angles between and the , , and global directions respectivelyˆ, , angles between and the , , and global directions respectively
ˆ, , angles between
x y z
x y z
x y z
i i j k
j i j k
θ θ θ
ψ ψ ψ
φ φ φ
≡
≡
≡ and the , , and global directions respectivelyk i j k
MECH 420: Finite Element Applications
Lecture 12: Grid Elements and Grid Analysis.
For grid analysis, orientation changes are only about the y direction.
Angle of rotation from the global to the elemental
2π θ−
zz
ˆ, y y
Plan View
ˆˆ ˆ( ) ( ) ( ) ˆˆˆ ˆ ˆ( ) ( ) ( )ˆ ˆˆ ˆ( ) ( ) ( )
i i j i k ix xy i j j j k j yz zi k j k k k
⎡ ⎤⋅ ⋅ ⋅⎧ ⎫ ⎧ ⎫⎢ ⎥⎪ ⎪ ⎪ ⎪= ⋅ ⋅ ⋅⎢ ⎥⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⋅ ⋅ ⋅⎩ ⎭ ⎩ ⎭⎢ ⎥⎣ ⎦
ˆ ˆ( ) ( ) 0ˆ ˆ( ) ( ) 0ˆ( ) 1
ˆ( ) cos( ) cos( )cos( ) sin( )sin( ) sin( )2 2 2
ˆ( ) cos( ) sin( )2
ˆ( ) cos( )ˆ( ) cos( )
i j j i
k j j k
j j
i k
k i
i i
k k
π π πθ θ θ θ
π θ θ
θ
θ
⋅ = ⋅ =
⋅ = ⋅ =
⋅ =
⋅ = − − = − − = −
⋅ = − = +
⋅ =
⋅ =
Sense of θ defined by RHR (y-axis).
MECH 420: Finite Element Applications
Lecture 12: Grid Elements and Grid Analysis.
For rotations θx, θy, or θz about the X, Y, or Z axes, respectively, of the global reference frame, the elemental coordinates are related to the global coordinates by:
ˆ1 0 0ˆ0ˆ0
x xy c s yz s c z
θ θθ θ
⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥= −⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎩ ⎭
ˆ0ˆ0 1 0ˆ0
x c s xy yz s c z
θ θ
θ θ
⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎩ ⎭
ˆ0ˆ0ˆ0 0 1
x c s xy s c yz z
θ θθ θ
−⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎩ ⎭
Rotation about only the x axis.
Rotation about only the y axis.
Rotation about only the z axis.
MECH 420: Finite Element Applications
Lecture 12: Grid Elements and Grid Analysis.
The rotation matrix applied in grid analysis is…
Consider the vector quantities that are being transformed from global to elemental coordinates.At node 1:
ˆ0ˆ0 1 0ˆ0
x c s xy yz s c z
θ θ
θ θ
⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎩ ⎭
11
1 1
1 1
ˆ0ˆ0 1 0ˆ0
xx
y y
z z
ff c sf ff s c f
θ θ
θ θ
⎧ ⎫⎧ ⎫ ⎡ ⎤ ⎪ ⎪⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎪ ⎪⎩ ⎭
In grid analysis the vertical loads are aligned with the global y axis. There is no need to transform the transverse forces or the corresponding displacements.
0.0
0.0
MECH 420: Finite Element Applications
Lecture 12: Grid Elements and Grid Analysis.
Consider the moments applied at the grid element node points.At node 1:
Applying this transformation at both nodes for the element load and displacement vectors…
1 1
1 1
1 1
ˆ0ˆ0 1 0ˆ0
x x
y y
z z
m c s mm mm s c m
θ θ
θ θ
⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥=⎨ ⎬ ⎨ ⎬⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥−⎩ ⎭ ⎣ ⎦ ⎩ ⎭
0.0
1 1
1 1
ˆˆ
x x
z z
m mc sm ms c
θ θθ θ
⎧ ⎫ ⎧ ⎫⎡ ⎤=⎨ ⎬ ⎨ ⎬⎢ ⎥−⎣ ⎦⎩ ⎭ ⎩ ⎭
MECH 420: Finite Element Applications
Lecture 12: Grid Elements and Grid Analysis.
3 2 3 2
1
1
2 21
23 2 3 2
2
2
2 2
12 0 6 12 0 6
0 0 0 0
6 0 4 6 0 2
12 0 6 12 0 6
0 0 0 0
6 0 2 6 0 4
y
x
zG
y
x
z
EI EI EI EIL L L L
GJ GJf L L
m EI EI EI EIm L L L LTf EI EI EI EI
L L L LmGJ GJmL L
EI EI EI EIL L L L
⎡ ⎤−⎢ ⎥⎢ ⎥⎢ ⎥−⎧ ⎫⎧ ⎫ ⎢ ⎥⎪ ⎪⎪ ⎪ ⎢ ⎥⎪ ⎪⎪ ⎪ −⎢ ⎥⎪ ⎪⎪ ⎪ ⎢ ⎥=⎨ ⎨ ⎬⎬ ⎢ ⎥⎪ ⎪ ⎪⎪ − − −⎢ ⎥⎪ ⎪ ⎪⎪ ⎢⎪ ⎪ ⎪⎪ ⎢⎩ ⎭ −⎩ ⎭ ⎢⎢⎢ −⎣ ⎦
1
1
1
2
2
2
1 0 0 0 0 00 0 0 00 0 0 0
; 0 0 0 1 0 00 0 0 00 0 0 0
y
x
zG G
y
x
z
dc ss c
T Td
c ss c
φ θ θφ θ θ
θ θφθ θφ
⎧ ⎫⎧ ⎫ ⎡ ⎤⎪ ⎪⎪ ⎪ ⎢ ⎥−⎪ ⎪⎪ ⎪ ⎢ ⎥⎪ ⎪⎪ ⎪ ⎢ ⎥=⎨ ⎨ ⎬⎬ ⎢ ⎥⎪ ⎪ ⎪⎪ ⎢ ⎥⎪ ⎪ ⎪⎪ −⎢ ⎥⎥ ⎪ ⎪ ⎪⎪ ⎢ ⎥⎥ ⎣ ⎦⎩ ⎭⎩ ⎭⎥⎥⎥
k
df
ˆ TG Gf T k T d⎡ ⎤= ⎣ ⎦
k
MECH 420: Finite Element Applications
Lecture 12: Grid Elements and Grid Analysis.
A note on “transforming rotations.”Sequences of finite rotation angles are not vectors.
2xπφ =
x
y
z
2zπφ = −
x
y
z
and then z xφ φ
x
y
z
and then x zφ φ
a vector prior to a sequence of active rotations.the same vector after the rotations.
rr≡′ ≡
r ′
r
r ′
MECH 420: Finite Element Applications
Lecture 12: Grid Elements and Grid Analysis.
Vectors are a superposition of scalar quantities that are each associated with a direction. Rotation sequences are NOT commutative and thus can’t be vectors.
Rotations must be expressed using rotation matrices. Examples are the 3x3 matrices we have looked at in this course.
x z z xi k k iφ φ φ φ+ ≠ + Rotation vectors DO NOT exist.
MECH 420: Finite Element Applications
Lecture 12: Grid Elements and Grid Analysis.
But if the two angles were both very small…
and then :0 1 0 00 0
0 0 1 0Note:
1 0 0 00 00 0 0 1
z x
z z
z z x x
x x
z z
x x z z
x x
x c s xy s c c s yz s c z
x c s xy c s s c yz s c z
φ φφ φφ φ φ φ
φ φ
φ φφ φ φ φφ φ
′ −⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥′ = −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪′ ⎢ ⎥ ⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎣ ⎦ ⎩ ⎭
′ −⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥′ ≠ −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪′ ⎢ ⎥ ⎢ ⎥⎩ ⎭ ⎣ ⎦ ⎣ ⎦ ⎩ ⎭
Can’t change the order of matrix multiplication.
MECH 420: Finite Element Applications
Lecture 12: Grid Elements and Grid Analysis.
0 1 0 0 1 0 0 00 0 0
a
00 0
nd
1 0 0 0 0 1
:
1 0
z z z z
z z x x x x z z
x
z
x x
z
z
x
x x
z
cd sd x cd sd xsd cd cd sd y cd sd sd cd y
sd cd z sd cd z
d dxyz
dd
φ φ φ φφ
φ φ φ φ
φ φ φ φ φ φ φφ φ φ
φφ
φ
→ →
′⎧ ⎫ − −⎡ ⎤ ⎡ ⎤ ⎧ ⎫ ⎡ ⎤ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥− = −⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎩ ⎭ ⎣ ⎦ ⎣ ⎦ ⎩ ⎭
⎪ ⎪′ =⎨ ⎬⎪ ⎪′⎩ ⎭
= −0 0
1 00 1 0 0
0
z
x z x
x x
x
z
x x d xd y y d d y
d z z d z
x d xy yz d z
φφ φ φ
φ φ
φ
φ
⎡ ⎤ ⎧ ⎫ ⎧ ⎫ ⎡ ⎤ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥= + −⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢ ⎥ ⎢ ⎥− −⎣ ⎦ ⎩ ⎭ ⎩ ⎭ ⎣ ⎦ ⎩ ⎭
⎧ ⎫ ⎧ ⎫ ⎧ ⎫⎪ ⎪ ⎪ ⎪ ⎪ ⎪= + ×⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭ ⎩ ⎭
The rotations would form a vector entity. But this is only strictly true in the case of infinitesimal rotations
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
§5.2 Rigid Plane Frame Examples.Pg.# 192-210 contains Examples 5.1 through 5.4.Here we look at P.5.3 pg.#241.
Find the nodal displacements at node 2 and the reaction forces at node 1. Draw the V and M diagrams for element #1. Select a channel section that ensures the bending stress is 66% of the yield stress for A36 steel.
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
Before embarking on a long solution process – target the necessary calculations.Only concerned with node 1 reactions and node 2 deflections.Problem has a few stages:
Solve for the deflections at node 2.Recover the element nodal loads on element 1 (including reactions at node 1).Using the element nodal loads draw the V and M diagrams (Vconstant and M linear).Based on the peak M value and allowable bending stress, size a cross section.
Involves some criterion on the allowable stress levels and some tabulated cross section geometries.
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
Could carry the channel geometry through as a variable or use guess and check…
x
y
z
Neutral axis of channel
4
1.920 in6.000 in13.1 inz
bhI
===
C6×8.2 American Standard Channel
MECH 420: Finite Element Applications
Element #1:
The 1st three equations will be discarded until the recovery stage and only the bottom 3x3 submatrix will be used in the assembly.
Lecture 12: Frame Analysis Example.
(1)1 3 2 3 2(1)
1(1) 2 21
(1)2
(1)2(1)
3 2 3 22
2 2
0 0 0 0
ˆ 0 12 6 0 12 6ˆ
0 6 4 0 6 2ˆˆ
0 0 0 0ˆ
0 12 6 0 12 6ˆ
0 6 2 0 6 4
x
y
z
x
y
z
AE AEL L
EI EI EI EIfL L L L
f EI EI EI EIm L L L L
AE AEfL Lf
EI EI EI EIm L L L L
EI EI EI EIL L L L
⎡ ⎤−⎢ ⎥⎢⎢⎧ ⎫ −⎢⎪ ⎪⎢⎪ ⎪
−⎢⎪ ⎪⎪ ⎪ ⎢=⎨ ⎬ ⎢⎪ ⎪ −⎢⎪ ⎪ ⎢⎪ ⎪ ⎢ − − −⎪ ⎪⎩ ⎭ ⎢⎢⎢ −⎣ ⎦
1
1
1
2
2
2
ˆˆ
ˆˆˆ
ˆ
x
y
z
x
y
z
dd
dd
φ
φ
⎥⎧ ⎫⎥⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥ ⎪ ⎪⎥ ⎨ ⎬⎥ ⎪ ⎪⎥ ⎪ ⎪⎥ ⎪ ⎪⎥ ⎪ ⎪⎥ ⎩ ⎭⎥⎥
(1)
(1)
1.00.0
CS
=
=0.0
0.0
0.0
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
Element #2:
The full 6x6 element #2 equations need to be filled out.
2 2 2 22 2 2 2
(2)2 2 2 22
2 2 2(2)
2(2)2
(2)3
(2)3(2)3
12 12 6 12 12 6
ˆ 12 6 12 12 6ˆ
6 6ˆ 4ˆ
ˆ
ˆ
x
y
z
x
y
z
I I I I I IAC S A CS S AC S A CS SL LL L L L
I I I I If AS C C A CS AS C CL LL L Lf
I Im E I SL LLf
fm
⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ − − − + − − −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠
⎧ ⎫ ⎛ ⎞ ⎛ ⎞+ − − − +⎪ ⎪ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎪ ⎪
⎪ ⎪⎪ ⎪ −=⎨ ⎬⎪ ⎪⎪ ⎪⎪ ⎪⎪ ⎪⎩ ⎭
2
2
2
32 2
2 23
2 2 32
ˆ
ˆ
ˆ2ˆ
12 12 6ˆ
ˆ12 6
SYM 4
x
y
z
x
y
z
d
d
C IdI I IAC S A CS S dLL L
I IAS C CLLI
φ
φ
⎡ ⎤⎢ ⎥⎢ ⎥
⎧ ⎫⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥⎪ ⎪⎢ ⎥ ⎨ ⎬⎢ ⎥ ⎪ ⎪⎢ ⎥⎛ ⎞ ⎪ ⎪+ −⎜ ⎟⎢ ⎥ ⎪ ⎪⎝ ⎠⎢ ⎥ ⎪ ⎪⎢ ⎥ ⎩ ⎭+ −⎢ ⎥
⎢ ⎥⎣ ⎦
(2)
(2)
0.7070.707
CS
=
=
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
Element #3:
The last 3 equations are being discarded in this particular problem (homogeneous conditions at Node ‘3’). Only the top 3x3 submatrix needs be computed.
(3)3 3 2 3 2(3)
3(3) 2 23
(3)4
(3)4(3)
3 2 3 24
2 2
0 0 0 0
ˆ 0 12 6 0 12 6ˆ
0 6 4 0 6 2ˆˆ
0 0 0 0ˆ
0 12 6 0 12 6ˆ
0 6 2 0 6 4
x
y
z
x
y
z
AE AEL L
EI EI EI EIfL L L L
f EI EI EI EIm L L L L
AE AEfL Lf
EI EI EI EIm L L L L
EI EI EI EIL L L L
⎡ ⎤−⎢ ⎥⎢⎢⎧ ⎫ −⎢⎪ ⎪⎢⎪ ⎪
−⎢⎪ ⎪⎪ ⎪ ⎢=⎨ ⎬ ⎢⎪ ⎪ −⎢⎪ ⎪ ⎢⎪ ⎪ ⎢ − − −⎪ ⎪⎩ ⎭ ⎢⎢⎢ −⎣ ⎦
3
3
3
4
4
4
ˆˆ
ˆˆˆ
ˆ
x
y
z
x
y
z
dd
dd
φ
φ
⎥⎧ ⎫⎥⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥ ⎪ ⎪⎥ ⎨ ⎬⎥ ⎪ ⎪⎥ ⎪ ⎪⎥ ⎪ ⎪⎥ ⎪ ⎪⎥ ⎩ ⎭⎥⎥
(3)
(3)
1.00.0
CS
=
=
0.0
0.0
0.0
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
The layout of the assembled system is…
(1) (2)2 22(1) (2)
2 22(1) (2)2 22
(2) (3)3 3 3
(2) (3)33 3(2) (3)23 3
ˆ ˆ0ˆ ˆ2000ˆ ˆ0ˆ ˆ0
2000 ˆ ˆ0 ˆ ˆ
x xx
y yy
z zz
x x x
yy y
zz z
f fFf fFm mm
F f fF f fm m m
⎧ ⎫+ ⎡ ⎤⎧ ⎫ ⎧ ⎫ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ +− ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎢ ⎥⎪ ⎪ ⎪ ⎪ +⎪ ⎪ ⎪ ⎪ ⎣ ⎦= = =⎨ ⎬ ⎨ ⎬ ⎨ ⎬⎡+⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎢⎪ ⎪ ⎪ ⎪ ⎪ ⎪− +⎪ ⎪ ⎪ ⎪ ⎪ ⎪
⎪ ⎪ ⎩ ⎭ ⎪ ⎪⎩ ⎭ ⎣+⎩ ⎭
2
2
2
3
3
3
ˆ
ˆ
ˆ
ˆ
ˆ
ˆ
x
y
z
x
y
z
d
d
d
d
φ
φ
⎧ ⎫⎡ ⎤⎡ ⎤ ⎪ ⎪⎢ ⎥⎢ ⎥ ⎪ ⎪⎢ ⎥⎢ ⎥ ⎪ ⎪⎢ ⎥⎢ ⎥ ⎪ ⎪⎢ ⎥⎢ ⎥ ⎨ ⎬⎢ ⎥⎤⎢ ⎥ ⎪ ⎪⎢ ⎥⎥⎢ ⎥ ⎪ ⎪⎢ ⎥⎢ ⎥⎢ ⎥ ⎪ ⎪⎢ ⎥⎢ ⎥⎢ ⎥ ⎪ ⎪⎦⎣ ⎦⎣ ⎦ ⎩ ⎭
Element #3
Element #1 Element #2
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
Calculating the physical and geometric system values:
Element #1:
Element #2:
Element #3:
Elements #1 and #3 should have identical element equations.
x
2ˆ 24 ksi3
ˆ is the distance from N.A. to surface; is the 2nd moment of area of the section.
x Y
zz
Mc c II
σ σ
σ
< =
= →
2 3 52 3
12 6 lbf0.03032 in ; 1.092 in ; 4.028 10 in
z zI I EL L L
= = = ×
2 3 52 3
12 6 lbf0.008528 in ; 0.5789 in ; 2.136 10 in
z zI I EL L L
= = = ×
2 3 52 3
12 6 lbf0.03032 in ; 1.092 in ; 4.028 10 in
z zI I EL L L
= = = ×
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
The assembly process leaves us with (imposing the BC’s simultaneously):
1
1
51
2
2
2
0 12.24 2.554 0.8745 2.573 2.554 0.87452000 2.695 3.523 2.554 2.573 0.87450 323.0 0.8745 0.8745 55.97
100 12.24 2.55 0.8745
2000 2.69 3.5230 323.0
x
y
z
x
y
z
FFMFFM
⎧ ⎫ − − − −⎧ ⎫ ⎡⎪ ⎪ ⎪ ⎪ ⎢− − − −⎪ ⎪ ⎪ ⎪ ⎢⎪ ⎪ −⎪ ⎪ ⎢= =⎨ ⎬ ⎨ ⎬ ⎢⎪ ⎪ ⎪ ⎪ ⎢⎪ ⎪ ⎪ ⎪− ⎢⎪ ⎪ ⎪ ⎪ ⎢⎩ ⎭ ⎣⎩ ⎭ i
2
2
2
3
3
3
x
y
z
x
y
z
dd
dd
φ
φ
⎧ ⎫⎤⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥⎨ ⎬⎥⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥⎦ ⎩ ⎭
92
2
29
3
3
3
3.01 10 in0.402 in
0.00666 rad3.30 10 in
0.402 in0.00666 rad
x
y
z
x
y
z
dd
dd
φ
φ
−
−
⎧ ⎫ ⎧− × ⎫⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪⎪ ⎪ −⎪ ⎪=⎨ ⎬ ⎨ ⎬×⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪−⎪ ⎪ ⎪ ⎪
⎩ ⎭⎩ ⎭
Note the magnitude of the rotations.
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
Step 7: Recovery.Start with the elemental nodal loads.Pull out element #1 and consider its equilibrium:
(1)11
(1)11
(1)1 15(1)
2 2
(1)2 2(1)2 2
ˆˆ2.4 0 0
ˆˆ 0 0.0303 1.0917ˆˆ 0 1.0917 26.2
4.025 10ˆ ˆˆ ˆ
ˆˆ
xx
yy
z z
x x
y y
z z
dfdf
mf df dm
φ
φ
⎧ ⎫⎧ ⎫ −⎡ ⎤ ⎪ ⎪⎪ ⎪⎢ ⎥ ⎪ ⎪⎪ ⎪⎢ ⎥ ⎪ ⎪⎪ ⎪ −⎪ ⎪ ⎪ ⎪⎢ ⎥= ×⎨ ⎬ ⎨ ⎬⎢ ⎥
⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎪ ⎪ ⎪ ⎪⎢ ⎥⎣ ⎦⎪ ⎪ ⎪ ⎪⎩ ⎭ ⎩ ⎭
0.0
0.0
0.0
(1)1
(1)1
3(1)1
ˆ 0.0 lbfˆ 2000 lbf
106.9 10 lbf inˆ
x
y
z
ffm
⎧ ⎫ ⎧ ⎫⎪ ⎪⎪ ⎪ ⎪ ⎪=⎨ ⎬ ⎨ ⎬⎪ ⎪ ⎪ ⎪× ⋅⎩ ⎭⎪ ⎪⎩ ⎭
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
The remaining elemental node 2 loads can be obtained by an FBD (the element must be in static equilibrium).
x
y
(1)1ˆ 106,900 lbf inzm = ⋅
(1)1 2000 lbfyf = (1)
2 2000 lbfyf = −
( )(1) (1) (1)2 1
ˆ 0 :ˆ ˆ2000 37060 lbf in
z
z z
mm L m
=
∴ = − = ⋅∑
(1)2ˆ 37,060 lbf inzm = ⋅
Note: no portion of the external 2000 lbf load is applied to element #2.
1
1 2
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
For element #2 it only remains to calculate the elemental nodal moments.
Using the element #2 equations in terms of global components…(2) (2)2 2(2) (2)3 3
(2)2(2)
2(2)
52(2)
3(2)
3(2)3
ˆ
ˆ
ˆ 0.4094 0.4094 52.4 0.4094 0.4094 26.22.136 10
ˆ 0.4094 0.4094 26.2 0.4094 0.4094 52.4
z z
z z
x
y
z
x
y
z
m m
m m
ffmffm
=
=
⎧ ⎫ ⎡ ⎤⎪ ⎪ ⎢⎪ ⎪ ⎢⎪ ⎪ ⎢− −⎪ ⎪ = ×⎨ ⎬ ⎢⎪ ⎪ ⎢⎪ ⎪ ⎢⎪ ⎪ ⎢
− −⎪ ⎪ ⎣⎩ ⎭
… … … … … …… … … … … …
… … … … … …… … … … … …
2
2
2
3
3
3
x
y
z
x
y
z
dd
dd
φ
φ
⎧ ⎫⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥ ⎪ ⎪⎨ ⎬⎥⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎥⎪ ⎪⎦ ⎩ ⎭
(2)2(2)3
ˆ 37,060 lbf inˆ 37,060 lbf in
z
z
m
m
= − ⋅
= ⋅Was there any need to
even evaluate these values? Could you have
immediately deduced them from the previous
slide?
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
The FBD for element #2 becomes…
x
y
(2)2ˆ 37,060 lbf inzm = − ⋅
(2)2 0.0 lbfyf = (2)
3 0.0 lbfyf =
(2)3ˆ 37,060 lbf inzm = ⋅
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
Recalling that:
The FBD of element #3 can be drawn right away:
(2) (3)3 3 3y y yF f f= +
x
y
(3)2ˆ 37,060 lbf inzm = − ⋅
(3)3 2000 lbfyf = − (3)
3 2000 lbfyf = +
(3)3ˆ 106,900 lbf inzm = − ⋅
( )(3) (1) (1)3 1
ˆ 0 :ˆ ˆ2000 106,900 lbf in
z
z z
mm L m
=
∴ = − − = − ⋅∑
MECH 420: Finite Element Applications
Lecture 12: Frame Analysis Example.
The peak bending moment in the system (neglecting any concentrated bend effects at the interconnections) is 106,900 lbfin.
x
y
z
Neutral axis of channel
3106,900 1 6.00 24.5 10 psi13.1 2
MAXMAXx
z
M cI
σ = = ⋅ = ×
4
1.920 in6.000 in13.1 inz
bhI
===
Made a good choice.
Note: we have only considered bending stresses – we assumed the transverse loads were applied through the shear centre, ‘SC’.